Problem stringlengths 5 967 | Rationale stringlengths 1 2.74k | options stringlengths 37 164 | correct stringclasses 5 values | annotated_formula stringlengths 7 1.65k | linear_formula stringlengths 8 925 | category stringclasses 6 values | answer stringclasses 5 values |
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the average weight of a group of boys is 20 kg . after a boy of weight 31 kg joins the group , the average weight of the group goes up by 1 kg . find the number of boys in the group originally ? | "let the number off boys in the group originally be x . total weight of the boys = 20 x after the boy weighing 31 kg joins the group , total weight of boys = 20 x + 31 so 20 x + 31 = 21 ( x + 1 ) = > x = 10 . answer : e" | a ) 12 , b ) 14 , c ) 18 , d ) 24 , e ) 10 | e | add(subtract(31, add(20, 1)), 1) | add(n0,n2)|subtract(n1,#0)|add(#1,n2)| | general | E |
16 people can write 52 book in 12 days working 8 hour a day . then in how many day 206 can be written by 64 people ? | "work per day epr hour per person = 52 / ( 12 * 8 * 16 ) / / eq - 1 people = 64 ; let suppose day = p ; per day work for 8 hours acc . to condition work per day epr hour per person = 206 / ( p * 8 * 64 ) / / eq - 2 eq - 1 = = eq - 2 ; p = 309 / 26 answer : a" | a ) 309 / 26 , b ) 309 / 28 , c ) 309 / 22 , d ) 319 / 26 , e ) 339 / 26 | a | divide(multiply(52, 12), divide(64, const_2)) | divide(n5,const_2)|multiply(n1,n2)|divide(#1,#0)| | physics | A |
m and n are the x and y coordinates , respectively , of a point in the coordinate plane . if the points ( m , n ) and ( m + p , n + 9 ) both lie on the line defined by the equation x = ( y / 3 ) - ( 2 / 5 ) , what is the value of p ? | "x = ( y / 3 ) - ( 2 / 5 ) , and so y = 3 x + 6 / 5 . the slope is 3 . ( n + 9 - n ) / ( m + p - m ) = 3 p = 3 the answer is c ." | a ) 1 , b ) 2 , c ) 3 , d ) 4 , e ) 5 | c | divide(9, 3) | divide(n0,n1)| | general | C |
in 1990 the budgets for projects q and v were $ 540,000 and $ 780,000 , respectively . in each of the next 10 years , the budget for q was increased by $ 30,000 and the budget for v was decreased by $ 10,000 . in which year was the budget for q equal to the budget for v ? | let the no of years it takes is x . 540 + 30 x = 780 - 10 x - - > 40 x = 240 and x = 6 . thus , it happens in 1996 . e . | a ) 1992 , b ) 1993 , c ) 1994 , d ) 1995 , e ) 1996 | e | add(1990, multiply(10, multiply(const_2, const_3))) | multiply(const_2,const_3)|multiply(n3,#0)|add(n0,#1) | general | E |
how many multiples of 5 are there between 0 and 358 ? | "5 * 1 = 5 5 * 71 = 355 total number of multiples = ( 71 + 1 ) = 72 answer d" | a ) 54 , b ) 75 , c ) 76 , d ) 71 , e ) 58 | d | add(divide(subtract(358, 0), 5), const_1) | subtract(n2,n1)|divide(#0,n0)|add(#1,const_1)| | general | D |
by weight , liquid x makes up 0.8 percent of solution a and 1.8 percent of solution b . if 300 grams of solution a are mixed with 700 grams of solution b , then liquid x accounts for what percent of the weight of the resulting solution ? | "i think there is a typo in question . it should have been ` ` by weight liquid ' x ' makes up . . . . . ` ` weight of liquid x = 0.8 % of weight of a + 1.8 % of weight of b when 300 gms of a and 700 gms of b is mixed : weight of liquid x = ( 0.8 * 300 ) / 100 + ( 1.8 * 700 ) / 100 = 15 gms % of liquid x in resultant mixture = ( 15 / 1000 ) * 100 = 1.5 % answer : 1.5 % answer : a" | a ) 1.5 % , b ) 1.9 % , c ) 10 % , d ) 15 % , e ) 19 % | a | divide(add(multiply(300, 0.8), multiply(700, 1.8)), const_1000) | multiply(n0,n2)|multiply(n1,n3)|add(#0,#1)|divide(#2,const_1000)| | gain | A |
a train 125 m long passes a man , running at 8 km / hr in the same direction in which the train is going , in 10 seconds . the speed of the train is : | speed of the train relative to man = ( 125 / 10 ) m / sec = ( 25 / 2 ) m / sec . [ ( 25 / 2 ) * ( 18 / 5 ) ] km / hr = 45 km / hr . let the speed of the train be x km / hr . then , relative speed = ( x - 8 ) km / hr . x - 8 = 45 = = > x = 53 km / hr . answer : a | a ) 53 , b ) 50 , c ) 99 , d ) 288 , e ) 12 | a | divide(divide(subtract(125, multiply(multiply(8, const_0_2778), 8)), 8), const_0_2778) | multiply(n1,const_0_2778)|multiply(n1,#0)|subtract(n0,#1)|divide(#2,n1)|divide(#3,const_0_2778)| | physics | A |
if x ^ 2 + ( 1 / x ^ 2 ) = 5 , x ^ 4 + ( 1 / x ^ 4 ) = ? | "- > x ^ 4 + ( 1 / x ^ 4 ) = ( x ^ 2 ) ^ 2 + ( 1 / x ^ 2 ) ^ 2 = ( x ^ 2 + 1 / x ^ 2 ) ^ 2 - 2 x ^ 2 ( 1 / x ^ 2 ) = 5 ^ 2 - 2 = 23 . thus , the answer is a ." | a ) 23 , b ) 11 , c ) 12 , d ) 14 , e ) 15 | a | subtract(power(2, 2), 2) | power(n3,n0)|subtract(#0,n0)| | general | A |
a sum of money at simple interest amounts to rs . 830 in 3 years and to rs . 854 in 4 years . the sum is : | "s . i . for 1 year = rs . ( 854 - 830 ) = rs . 24 . s . i . for 3 years = rs . ( 24 x 3 ) = rs . 72 . principal = rs . ( 830 - 72 ) = rs . 758 . answer : c" | a ) 647 , b ) 698 , c ) 758 , d ) 847 , e ) 976 | c | subtract(830, divide(multiply(subtract(854, 830), 3), 4)) | subtract(n2,n0)|multiply(n1,#0)|divide(#1,n3)|subtract(n0,#2)| | gain | C |
the average of 11 results is 20 . the average of first 5 of them is 15 and that of last 5 is 22 . find the 6 th result ? | "6 th result = sum of 11 results - sum of 10 results = 11 * 20 - 5 * 15 - 5 * 22 = 220 - 75 - 110 = 35 answer is a" | a ) 35 , b ) 50 , c ) 100 , d ) 120 , e ) 150 | a | subtract(subtract(multiply(11, 20), multiply(5, 22)), multiply(5, 15)) | multiply(n0,n1)|multiply(n2,n5)|multiply(n2,n3)|subtract(#0,#1)|subtract(#3,#2)| | general | A |
what will be the fraction of 25 % | "explanation : it will 25 * 1 / 100 = 1 / 4 option a" | a ) 1 / 4 , b ) 1 / 5 , c ) 1 / 10 , d ) 1 / 11 , e ) none of above | a | divide(circle_area(divide(25, const_2)), const_2) | divide(n0,const_2)|circle_area(#0)|divide(#1,const_2)| | gain | A |
the maximum number of girls among them 2923 bags and 3239 eyeliners can be distributed in such a way that each student gets the same number of bags and eyeliners . | max no of girls = hcf of 2923 and 3239 = 79 answer : b | a ) 80 , b ) 79 , c ) 78 , d ) 81 , e ) 82 | b | gcd(2923, 3239) | gcd(n0,n1) | general | B |
90 people are attending a newspaper conference . 45 of them are writers and more than 38 are editors . of the people at the conference , x are both writers and editors and 2 x are neither . what is the largest possible number of people who are both writers and editors ? | { total } = { writers } + { editors } - { both } + { neither } . { total } = 90 ; { writers } = 45 ; { editors } > 38 ; { both } = x ; { neither } = 2 x ; 90 = 45 + { editors } - x + 2 x - - > x = 45 - { editors } . we want to maximize x , thus we should minimize { editors } , minimum possible value of { editors } is 39 , thus x = { both } = 45 - 39 = 6 . answer : e . | a ) 18 , b ) 16 , c ) 12 , d ) 10 , e ) 6 | e | subtract(90, add(add(38, const_1), 45)) | add(n2,const_1)|add(n1,#0)|subtract(n0,#1) | other | E |
a sells a bicycle to b at a profit of 30 % and b sells it to c at a loss of 20 % . find the resultant profit or loss . | the resultant profit or loss = 30 - 20 - ( 30 * 20 ) / 100 = 4 % profit = 4 % answer is b | a ) 5 % , b ) 4 % , c ) - 4 % , d ) - 12 % , e ) 2 % | b | multiply(subtract(multiply(add(const_1, divide(30, const_100)), subtract(const_1, divide(20, const_100))), const_1), const_100) | divide(n0,const_100)|divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#1)|multiply(#2,#3)|subtract(#4,const_1)|multiply(#5,const_100) | gain | B |
if xy = 9 , x / y = 36 , for positive numbers x and y , y = ? | very easy question . 2 variables and 2 easy equations . xy = 9 - - - > x = 9 / y - ( i ) x / y = 36 - - - > replacing ( i ) here - - - > 9 / ( y ^ 2 ) = 36 - - - > y ^ 2 = 9 / 36 = 1 / 4 - - - > y = 1 / 2 or - 1 / 2 the question states that x and y are positive integers . therefore , y = 1 / 2 is the answer . answer a . | a ) 1 / 2 , b ) 2 , c ) 1 / 3 , d ) 3 , e ) 1 / 6 | a | sqrt(divide(9, 36)) | divide(n0,n1)|sqrt(#0) | general | A |
a bag holds 2 red marbles and 3 green marbles . if you removed two randomly selected marbles from the bag , without replacement , what is the probability that both would be green ? | "given : 2 r and 3 g marbles required : probability that 2 marbles removed without replacement are both red initially we have to pick one red from a total of 2 red and 3 green marbles after one red has been picked , we need to pick 1 green from a total of 2 green and 2 red marbles . p ( both green ) = ( 3 / 5 ) * ( 2 / 4 ) = 3 / 10 option c" | a ) 1 / 10 , b ) 1 / 5 , c ) 3 / 10 , d ) 2 / 5 , e ) 1 / 2 | c | divide(choose(2, 2), choose(add(2, 3), 2)) | add(n0,n1)|choose(n0,n0)|choose(#0,n0)|divide(#1,#2)| | other | C |
a certain bus driver is paid a regular rate of $ 12 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 976 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 12 = 480 excess = 976 - 480 = 496 for extra hours = . 75 ( 12 ) = 9 + 12 = 21 number of extra hrs = 496 / 21 = 23.62 total hrs = 40 + 23.62 = 63.62 = 64 approx . answer d 64" | a ) 60 , b ) 68 , c ) 65 , d ) 64 , e ) 61 | d | add(40, divide(subtract(976, multiply(12, 40)), divide(multiply(12, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general | D |
if the cost price is 80 % of the selling price , then what is the profit percent ? | "let s . p . = $ 100 c . p . = $ 80 profit = $ 20 profit % = 20 / 80 * 100 = 25 / 6 = 25 % answer is d" | a ) 5 % , b ) 11 % , c ) 13 % , d ) 25 % , e ) 19 % | d | multiply(divide(subtract(const_100, 80), 80), const_100) | subtract(const_100,n0)|divide(#0,n0)|multiply(#1,const_100)| | gain | D |
a shopkeeper buys mangoes at the rate of 4 a rupee and sells them at 2 a rupee . find his net profit or loss percent ? | "the total number of mangoes bought by the shopkeeper be 12 . if he buys 4 a rupee , his cp = 3 he selling at 2 a rupee , his sp = 4 profit = sp - cp = 4 - 2 = 2 profit percent = 2 / 2 * 100 = 100 % answer : a" | a ) 100 % , b ) 200 % , c ) 250 % , d ) 300 % , e ) 50 % | a | divide(multiply(2, const_100), 4) | multiply(n1,const_100)|divide(#0,n0)| | gain | A |
find the average of all prime numbers between 80 and 100 | "prime numbers between 80 and 100 are 83 , 89 , 97 required average = ( 83 + 89 + 97 ) / 3 = 269 / 3 = 89.6 answer is d" | a ) 88 , b ) 60 , c ) 80 , d ) 89.6 , e ) 92 | d | divide(add(add(add(80, const_1), add(add(80, const_1), const_2)), add(subtract(100, 80), subtract(100, const_2))), 80) | add(n0,const_1)|subtract(n1,n0)|subtract(n1,const_2)|add(#0,const_2)|add(#1,#2)|add(#0,#3)|add(#5,#4)|divide(#6,n0)| | general | D |
water consists of hydrogen and oxygen , and the approximate ratio , by mass , of hydrogen to oxygen is 2 : 16 . approximately how many grams of oxygen are there in 180 grams of water ? | "solution : we are given that the ratio of hydrogen to oxygen in water , by mass , is 2 : 16 . using our ratio multiplier we can re - write this as 2 x : 16 x . we can now use these expressions to determine how much oxygen is 180 18 x = 180 x = 10 since x is 10 , we know that there are 16 x 10 = 160 grams of oxygen in 180 grams of water . answer a ." | a ) 160 , b ) 72 , c ) 112 , d ) 128 , e ) 142 | a | multiply(2, divide(180, add(2, 16))) | add(n0,n1)|divide(n2,#0)|multiply(n0,#1)| | other | A |
a rectangular grass field is 75 m * 55 m , it has a path of 3.2 m wide all round it on the outside . find the area of the path and the cost of constructing it at rs . 2 per sq m ? | "area = ( l + b + 2 d ) 2 d = ( 75 + 55 + 3.2 * 2 ) 2 * 2.5 = > 682 682 * 2 = rs . 1364 answer : d" | a ) s . 1350 , b ) s . 1327 , c ) s . 1328 , d ) s . 1364 , e ) s . 1927 | d | multiply(subtract(rectangle_area(add(75, multiply(3.2, 2)), add(55, multiply(3.2, 2))), rectangle_area(75, 55)), 2) | multiply(n2,n3)|rectangle_area(n0,n1)|add(n0,#0)|add(n1,#0)|rectangle_area(#2,#3)|subtract(#4,#1)|multiply(n3,#5)| | geometry | D |
3 candidates in an election and received 1136 , 8236 and 11628 votes respectively . what % of the total votes did the winning candidate gotin that election ? | "total number of votes polled = ( 1136 + 8236 + 11628 ) = 21000 so , required percentage = 11628 / 21000 * 100 = 55.4 % b" | a ) 40 % , b ) 55.4 % , c ) 57 % , d ) 60 % , e ) 62 % | b | multiply(divide(11628, add(add(1136, 8236), 11628)), const_100) | add(n1,n2)|add(n3,#0)|divide(n3,#1)|multiply(#2,const_100)| | gain | B |
a leak in the bottom of a tank can empty the tank in 6 hrs . an pipe fills water at the rate of 4 ltrs / minute . when the tank is full in inlet is opened and due to the leak the tank is empties in 8 hrs . the capacity of the tank is ? | "1 / x - 1 / 6 = - 1 / 8 x = 24 hrs 24 * 60 * 4 = 5760 e" | a ) 2345 , b ) 2350 , c ) 2457 , d ) 4657 , e ) 5760 | e | multiply(multiply(multiply(6, 4), const_60), 4) | multiply(n0,n1)|multiply(#0,const_60)|multiply(n1,#1)| | physics | E |
what is the smallest number which when increased by 3 is divisible by 9 , 70 , 25 and 21 ? | "when increased by 3 , the number must include at least 2 * 3 ^ 2 * 5 ^ 2 * 7 = 3150 the answer is b ." | a ) 2927 , b ) 3147 , c ) 3387 , d ) 3567 , e ) 3797 | b | add(lcm(lcm(9, 70), lcm(25, 21)), 3) | lcm(n1,n2)|lcm(n3,n4)|lcm(#0,#1)|add(n0,#2)| | general | B |
if there are 200 questions in a 3 hr examination . among these questions are 50 type a problems , which requires twice as much as time be spent than the rest of the type b problems . how many minutes should be spent on type a problems ? | "x = time for type b prolems 2 x = time for type a problem total time = 3 hrs = 180 min 150 x + 50 * 2 x = 180 x = 180 / 250 x = 0.72 time taken for type a problem = 50 * 2 * 0.72 = 72 min answer : a" | a ) 72 min , b ) 62 min , c ) 70 min , d ) 74 min , e ) 76 min | a | multiply(multiply(const_2, divide(multiply(3, const_60), add(subtract(200, 50), multiply(const_2, 50)))), 50) | multiply(n1,const_60)|multiply(n2,const_2)|subtract(n0,n2)|add(#1,#2)|divide(#0,#3)|multiply(#4,const_2)|multiply(n2,#5)| | general | A |
the greatest number by which the product of 3 consecutive multiples of 4 is always divisible is | solution required number = product of first three multiplies of 3 = ( 4 ã — 8 ã — 12 ) = 384 . answer d | a ) 54 , b ) 76 , c ) 152 , d ) 384 , e ) none of these | d | multiply(multiply(4, multiply(4, const_2)), multiply(4, 3)) | multiply(n1,const_2)|multiply(n0,n1)|multiply(n1,#0)|multiply(#2,#1) | general | D |
the ratio between the length and the breadth of a rectangular park is 3 : 2 . if a man cycling along the boundary of the park at the speed of 12 km / hr completes one round in 5 minutes , then the area of the park ( in sq . m ) is : | "perimeter = distance covered in 5 min . = ( 12000 / 60 ) x 5 m = 1000 m . let length = 3 x metres and breadth = 2 x metres . then , 2 ( 3 x + 2 x ) = 1000 or x = 100 . length = 300 m and breadth = 200 m . area = ( 300 x 200 ) m 2 = 60000 m 2 . answer : d" | a ) 153601 , b ) 153600 , c ) 153602 , d ) 60000 , e ) 153604 | d | rectangle_area(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 5), const_1000), add(3, 2)), const_2), multiply(divide(divide(multiply(multiply(divide(12, multiply(const_10, multiply(const_3, const_2))), 5), const_1000), add(3, 2)), const_2), 2)) | add(n0,n1)|multiply(const_2,const_3)|multiply(#1,const_10)|divide(n2,#2)|multiply(n3,#3)|multiply(#4,const_1000)|divide(#5,#0)|divide(#6,const_2)|multiply(n1,#7)|rectangle_area(#7,#8)| | physics | D |
while working alone at their constant rates , computer x can process 240 files in 12 hours , and computer y can process 240 files in 4 hours . if all files processed by these computers are the same size , how many hours would it take the two computers , working at the same time at their respective constant rates , to process a total of 240 files ? | "both computers together process files at a rate of 240 / 12 + 240 / 4 = 20 + 60 = 80 files per hour . the time required to process 240 files is 240 / 80 = 3 hours the answer is d ." | a ) 2 , b ) 2.4 , c ) 2.7 , d ) 3 , e ) 3.5 | d | divide(240, add(divide(240, 12), divide(240, 4))) | divide(n0,n1)|divide(n0,n3)|add(#0,#1)|divide(n0,#2)| | physics | D |
a rectangular floor is covered by a rug except for a strip 4 meters wide along each of the four edge . if the floor is 25 meters by 20 meters , what is the area of the rug in square meters ? | a strip of 4 meters is covering the inner rectangular rug for all 4 sides . length of inner rug = 25 - ( 2 * 4 ) breadth of inner rug = 20 - ( 2 * 4 ) area of rug = 17 * 12 = 204 sq . mt â nswer : c | ['a ) 256', 'b ) 266', 'c ) 204', 'd ) 224', 'e ) 324'] | c | rectangle_area(subtract(25, multiply(4, const_2)), subtract(20, multiply(4, const_2))) | multiply(n0,const_2)|subtract(n1,#0)|subtract(n2,#0)|rectangle_area(#1,#2) | geometry | C |
the mean daily profit made by a shopkeeper in a month of 30 days was rs . 350 . if the mean profit for the first fifteen days was rs . 265 , then the mean profit for the last 15 days would be | average would be : 350 = ( 265 + x ) / 2 on solving , x = 435 . answer : a | a ) rs . 435 , b ) rs . 350 , c ) rs . 275 , d ) rs . 425 , e ) none of these | a | divide(subtract(multiply(30, 350), multiply(15, 265)), 15) | multiply(n0,n1)|multiply(n2,n3)|subtract(#0,#1)|divide(#2,n3) | gain | A |
how many pieces of 75 cm can be cut from a rope 57 meters long ? | "explanation : total pieces of 75 cm that can be cut from a rope of 57 meters long is = ( 57 meters ) / ( 75 cm ) = ( 57 meters ) / ( 0.75 meters ) = 76 answer : c" | a ) 30 , b ) 40 , c ) 76 , d ) none , e ) can not be determined | c | divide(57, 75) | divide(n1,n0)| | physics | C |
dan ' s age after 16 years will be 4 times his age 8 years ago . what is the present age of dan ? | let dan ' s present age be x . x + 16 = 4 ( x - 8 ) 3 x = 48 x = 16 the answer is b . | a ) 12 , b ) 16 , c ) 20 , d ) 24 , e ) 28 | b | divide(add(16, multiply(8, 4)), subtract(4, const_1)) | multiply(n1,n2)|subtract(n1,const_1)|add(n0,#0)|divide(#2,#1) | general | B |
in a bus left side are 15 seats available , 3 few seats in right side because in rear exit door . each seat hold 3 people . in addition , there is a seat back can sit 9 people all together . how many people can sit in a bus ? | "right side = 15 seat left side = 15 - 3 ( 3 few seat in right side ) = 12 seat total = 15 + 12 = 27 people can seat in 27 seat = 27 * 3 = 81 people can seat in last seat = 9 total people can seat = 81 + 9 = 90 answer : d" | a ) 52 , b ) 49 , c ) 95 , d ) 90 , e ) 66 | d | add(multiply(add(15, subtract(15, 3)), 3), 9) | subtract(n0,n1)|add(n0,#0)|multiply(n1,#1)|add(n3,#2)| | general | D |
34 : 43 : : 25 : ? | "ans 52 reverse of 25 answer : b" | a ) 49 , b ) 52 , c ) 36 , d ) 64 , e ) 56 | b | multiply(25, divide(34, 43)) | divide(n0,n1)|multiply(n2,#0)| | general | B |
the average of 10 consecutive integers is 25 . then , 9 is deducted from the first consecutive number , 8 is deducted from the second , 7 is deducted form the third , and so on until the last number which remains unchanged . what is the new average ? | "the total subtracted is ( 9 + 8 + . . . + 1 ) = ( 9 * 10 ) / 2 = 45 on average , each number will be reduced by 45 / 10 = 4.5 therefore , the overall average will be reduced by 4.5 the answer is d ." | a ) 19 , b ) 19.5 , c ) 20 , d ) 20.5 , e ) 21 | d | divide(subtract(multiply(10, 25), multiply(add(const_4, const_1), 9)), 10) | add(const_1,const_4)|multiply(n0,n1)|multiply(n2,#0)|subtract(#1,#2)|divide(#3,n0)| | general | D |
the cost of an article was rs . 75 . the cost was first increased by 20 % and later on it was reduced by 20 % . the present cost of the article is : | solution : initial cost = rs . 75 after 20 % increase in the cost , it becomes , ( 75 + 20 % of 75 ) = rs . 90 now , cost is decreased by 20 % , so cost will become , ( 90 - 20 % of 90 ) = rs . 72 . so , present cost is rs . 72 . mind calculation method : 75 - - - - - 20 % ↑ - - → 90 - - - - - 20 % ↓ - - - - - → 72 . answer : option a | a ) rs . 72 , b ) rs . 60 , c ) rs . 75 , d ) rs . 76 , e ) none | a | multiply(multiply(75, add(const_1, divide(20, const_100))), subtract(const_1, divide(20, const_100))) | divide(n1,const_100)|add(#0,const_1)|subtract(const_1,#0)|multiply(n0,#1)|multiply(#3,#2) | gain | A |
of the 600 residents of clermontville , 35 % watch the television show island survival , 40 % watch lovelost lawyers and 50 % watch medical emergency . if all residents watch at least one of these 3 shows and 18 % watch exactly 2 of these shows , then how many clermontville residents z watch all of the shows ? | oa is d . 100 = a + b + c - ab - ac - bc + abc , which is the same as the following formula 100 = a + b + c + ( - ab - ac - bc + abc + abc + abc ) - 2 abc . the term between parantheses value 18 % so the equation to resolve is 100 = 35 + 40 + 50 - 18 - 2 abc therefore the value of abc is z = 3.5 % of 600 , is 21 . d is the correct answer | a ) 150 , b ) 108 , c ) 42 , d ) 21 , e ) - 21 | d | divide(subtract(subtract(add(add(multiply(600, divide(35, const_100)), multiply(600, divide(40, const_100))), multiply(600, divide(50, const_100))), 600), multiply(divide(18, const_100), 600)), 2) | divide(n1,const_100)|divide(n2,const_100)|divide(n3,const_100)|divide(n5,const_100)|multiply(n0,#0)|multiply(n0,#1)|multiply(n0,#2)|multiply(n0,#3)|add(#4,#5)|add(#8,#6)|subtract(#9,n0)|subtract(#10,#7)|divide(#11,n6) | gain | D |
tickets numbered from 1 to 27 are mixed and then a ticket is selected randomly . what is the probability that the selected ticket bears a number which is a multiple of 3 ? | "here , s = [ 1 , 2 , 3 , 4 , … . , 19 , 20 , 21 , 22 , 23 , 24 , 25 , 26 , 27 ] let e = event of getting a multiple of 3 = [ 3 , 6 , 9 , 12 , 15 , 18 , 21 , 24 , 27 ] p ( e ) = n ( e ) / n ( s ) = 9 / 27 = 1 / 3 the answer is a ." | a ) 1 / 3 , b ) 2 / 5 , c ) 3 / 10 , d ) 3 / 7 , e ) 1 / 7 | a | divide(divide(27, 3), 27) | divide(n1,n2)|divide(#0,n1)| | general | A |
by selling an article at rs . 400 , a profit of 60 % is made . find its cost price ? | "sp = 400 cp = ( sp ) * [ 100 / ( 100 + p ) ] = 400 * [ 100 / ( 100 + 60 ) ] = 400 * [ 100 / 160 ] = rs . 250 answer : d" | a ) 228 , b ) 267 , c ) 287 , d ) 250 , e ) 811 | d | divide(multiply(400, const_100), add(const_100, 60)) | add(n1,const_100)|multiply(n0,const_100)|divide(#1,#0)| | gain | D |
find the smallest number of five digits exactly divisible by 32 , 40,56 and 64 . | "smallest number of five digits is 10000 . required number must be divisible by l . c . m . of 32,40 , 56,64 i . e 2240 , on dividing 10000 by 2240 , we get 1200 as remainder . therefore , required number = 10000 + ( 2240 â € “ 1200 ) = 11040 . answer is c ." | a ) 11020 , b ) 11030 , c ) 11040 , d ) 11060 , e ) 11080 | c | add(subtract(multiply(const_10, multiply(const_100, const_100)), const_100), 40,56) | multiply(const_100,const_100)|multiply(#0,const_10)|subtract(#1,const_100)|add(n1,#2)| | general | C |
the average of 13 numbers is 59 . average of the first 7 of them is 57 and that of the last 7 is 61 . find the 8 th number ? | "sum of all the 13 numbers = 13 * 59 = 767 sum of the first 7 of them = 7 * 57 = 399 sum of the last 7 of them = 7 * 61 = 427 so , the 8 th number = 427 + 399 - 767 = 59 . answer : a" | a ) 59 , b ) 83 , c ) 45 , d ) 53 , e ) 64 | a | subtract(multiply(8, 61), subtract(multiply(13, 59), multiply(57, 7))) | multiply(n5,n6)|multiply(n0,n1)|multiply(n2,n3)|subtract(#1,#2)|subtract(#0,#3)| | general | A |
33 1 / 3 % of 330 ? | "33 1 / 3 % = 1 / 3 1 / 3 × 330 = 110 c )" | a ) 80 , b ) 90 , c ) 110 , d ) 120 , e ) 130 | c | divide(multiply(add(33, divide(1, 3)), 330), const_100) | divide(n1,n2)|add(n0,#0)|multiply(n3,#1)|divide(#2,const_100)| | gain | C |
how many 3 - digits number are palindromic numbers ? a palindromic number reads the same forward and backward , example 121 . | "take the task of building palindromes and break it intostages . stage 1 : select the hundreds digit we can choose 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 1 in 9 ways stage 2 : select the tens digit we can choose 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , or 9 so , we can complete stage 2 in 10 ways important : at this point , the remaining digits are alreadylocked in . stage 3 : select the units digit this digit must be the same as the thousands digit ( which we already chose in stage 2 ) so , we can complete this stage in 1 way . by thefundamental counting principle ( fcp ) , we can complete all 5 stages ( and thus build a 3 - digit palindrome ) in ( 9 ) ( 10 ) ( 1 ) ways ( = 900 ways ) answer : a" | a ) 90 , b ) 610 , c ) 729 , d ) 900 , e ) 1000 | a | multiply(multiply(multiply(multiply(3, const_10), const_10), const_10), const_10) | multiply(n0,const_10)|multiply(#0,const_10)|multiply(#1,const_10)|multiply(#2,const_10)| | general | A |
n is a positive integer . when n + 1 is divided by 4 , the remainder is 3 . what is the remainder when n is divided by 3 ? | "n + 1 = 4 a + 3 i . e . n + 1 = 7 , 10 , 13 , 16 , . . . etc . i . e . n = 6 , 9 , 12 , 15 , . . . etc . when n is divided by 3 remainder is always 0 answer : d" | a ) 1 , b ) 2 , c ) 3 , d ) 0 , e ) 4 | d | divide(4, add(3, 3)) | add(n2,n3)|divide(n1,#0)| | general | D |
find the average of all the numbers between 11 and 36 which are divisible by 5 . | "explanation : average = ( 15 + 20 + 25 + 30 + 35 ) / 5 = 125 / 5 = 25 answer : a" | a ) 25 , b ) 77 , c ) 20 , d ) 28 , e ) 10 | a | divide(add(add(11, const_4), subtract(36, const_4)), const_2) | add(n0,const_4)|subtract(n1,const_4)|add(#0,#1)|divide(#2,const_2)| | general | A |
a certain bus driver is paid a regular rate of $ 16 per hour for any number of hours that does not exceed 40 hours per week . for any overtime hours worked in excess of 40 hours per week , the bus driver is paid a rate that is 75 % higher than his regular rate . if last week the bus driver earned $ 752 in total compensation , how many total hours did he work that week ? | "for 40 hrs = 40 * 16 = 640 excess = 752 - 640 = 112 for extra hours = . 75 ( 16 ) = 12 + 16 = 28 number of extra hrs = 112 / 28 = 28 / 7 = 4 total hrs = 40 + 4 = 44 answer c 44" | a ) 36 , b ) 40 , c ) 44 , d ) 48 , e ) 52 | c | add(40, divide(subtract(752, multiply(16, 40)), divide(multiply(16, add(const_100, 75)), const_100))) | add(n3,const_100)|multiply(n0,n1)|multiply(n0,#0)|subtract(n4,#1)|divide(#2,const_100)|divide(#3,#4)|add(n1,#5)| | general | C |
in what time will a train 134 m long cross an electric pole , it its speed be 169 km / hr ? | "speed = 169 * 5 / 18 = 46.9 m / sec time taken = 134 / 46.9 = 2.9 sec . answer : c" | a ) 2.5 , b ) 2.7 , c ) 2.9 , d ) 2.3 , e ) 2.1 | c | divide(134, multiply(169, const_0_2778)) | multiply(n1,const_0_2778)|divide(n0,#0)| | physics | C |
a shopkeeper has 100 kg of apples . he sells 50 % of these at 25 % profit and remaining 50 % at 30 % profit . find his % profit on total . | "total number of apples = 100 let the cost price be x selling price at 25 % profit = 1.25 x selling price at 30 % profit = 1.3 x profit % = ( ( sp - cp ) / cp ) * 100 profit % = ( ( 1 / 2 ) * 100 * 1.25 x + ( 1 / 2 ) * 100 * 1.3 x - 100 x ) / 100 x * 100 = ( 255 - 200 ) / 2 = 27.5 % answer is a" | a ) 27.5 % , b ) 25.6 % , c ) 31.5 % , d ) 35.9 % , e ) 29.5 % | a | divide(multiply(subtract(add(multiply(divide(multiply(100, 50), const_100), divide(add(const_100, 25), const_100)), multiply(divide(multiply(100, 50), const_100), divide(add(const_100, 30), const_100))), 100), const_100), 100) | add(n2,const_100)|add(n4,const_100)|multiply(n0,n1)|multiply(n0,n3)|divide(#2,const_100)|divide(#0,const_100)|divide(#3,const_100)|divide(#1,const_100)|multiply(#4,#5)|multiply(#6,#7)|add(#8,#9)|subtract(#10,n0)|multiply(#11,const_100)|divide(#12,n0)| | gain | A |
there were two candidates in an election . winner candidate received 52 % of votes and won the election by 288 votes . find the number of votes casted to the winning candidate ? | "w = 52 % l = 48 % 52 % - 48 % = 4 % 4 % - - - - - - - - 288 52 % - - - - - - - - ? = > 3744 answer : b" | a ) 776 , b ) 3744 , c ) 299 , d ) 257 , e ) 125 | b | divide(multiply(divide(288, divide(subtract(52, subtract(const_100, 52)), const_100)), 52), const_100) | subtract(const_100,n0)|subtract(n0,#0)|divide(#1,const_100)|divide(n1,#2)|multiply(n0,#3)|divide(#4,const_100)| | gain | B |
kamal obtained 76 , 60 , 72 , 65 and 82 marks ( out of 100 ) in english , mathematics , physics , chemistry and biology . what are his average marks ? | "sol . average = 76 + 60 + 72 + 65 + 82 / 5 ) = ( 355 / 5 ) = 71 . answer a" | a ) 71 , b ) 69 , c ) 72 , d ) 75 , e ) none | a | divide(add(add(add(add(76, 60), 72), 65), 82), add(const_1, const_4)) | add(n0,n1)|add(const_1,const_4)|add(n2,#0)|add(n3,#2)|add(n4,#3)|divide(#4,#1)| | general | A |
a factory produces 4340 toys per week . if the workers at this factory work 2 days a week and if these workers make the same number of toys everyday , how many toys are produced each day ? | "to find the number of toys produced every day , we divide the total number of toys produced in one week ( of 2 days ) by 2 . 4340 / 2 = 2170 toys correct answer b" | a ) 1375 toys , b ) 2170 toys , c ) 2375 toys , d ) 2175 toys , e ) 5375 toys | b | divide(4340, 2) | divide(n0,n1)| | physics | B |
a cistern 9 m long and 4 m wide contains water up to a depth of 1 m 25 cm . the total area of the wet surface is : | "area of the wet surface = [ 2 ( lb + bh + lh ) - lb ] = 2 ( bh + lh ) + lb = [ 2 ( 4 x 1.25 + 9 x 1.25 ) + 9 x 4 ] m 2 = 68.5 m 2 . answer : option d" | a ) 49 m 2 , b ) 50 m 2 , c ) 53.5 m 2 , d ) 68.5 m 2 , e ) 57 m 2 | d | add(multiply(const_2, add(multiply(add(divide(25, const_100), 1), 4), multiply(add(divide(25, const_100), 1), 9))), multiply(4, 9)) | divide(n3,const_100)|multiply(n0,n1)|add(n2,#0)|multiply(n1,#2)|multiply(n0,#2)|add(#3,#4)|multiply(#5,const_2)|add(#6,#1)| | physics | D |
a number x is 11 times another number y . the percentage that y is less than x is | "say y = 1 and x = 11 . then y = 1 is less than x = 11 by ( 11 - 1 ) / 11 * 100 = 10 / 11 * 100 = 90.9 % . answer : b ." | a ) 12.5 % , b ) 90.9 , c ) 91.7 , d ) 11 % , e ) 1 % | b | multiply(divide(subtract(11, const_1), 11), const_100) | subtract(n0,const_1)|divide(#0,n0)|multiply(#1,const_100)| | general | B |
a store ’ s selling price of $ 2240 for a certain computer would yield a profit of 40 percent of the store ’ s cost for the computer . what selling price would yield a profit of 60 percent of the computer ’ s cost ? | "1.4 x = 2240 x = 2240 / 1.4 so , 1.6 x = 2240 * 1.6 / 1.4 = 2560 answer : - c" | a ) $ 2400 , b ) $ 2464 , c ) $ 2560 , d ) $ 2732 , e ) $ 2800 | c | multiply(2240, divide(add(const_100, 60), add(const_100, 40))) | add(n2,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(n0,#2)| | gain | C |
rose made a part payment of $ 300 toward buying a brand new car , which represents 5 % of the total cost of the car , how much remains to be paid ? | explanation : let ' s start with what the total price of the car would be . if 5 % is equal to $ 300 then 100 % equals $ x . we just have to multiply $ 300 by 20 to get total amount = $ 6000 . out of this amount we then need to deduct the amount already paid which was $ 300 so we have $ 6000 - $ 300 = $ 5700 answer : option a | a ) $ 5700 , b ) $ 5500 , c ) $ 5800 , d ) $ 5200 , e ) $ 5880 | a | subtract(divide(multiply(300, const_100), 5), 300) | multiply(n0,const_100)|divide(#0,n1)|subtract(#1,n0) | general | A |
an amount of money is to be divided between p , q and r in the ratio of 3 : 7 : 12 . if the difference between the shares of p and q is rs . 2400 , what will be the difference between q and r ' s share ? | "4 - - - 2000 5 - - - ? = > 3000 answer : b :" | a ) 1266 , b ) 3000 , c ) 2866 , d ) 2999 , e ) 2678 | b | multiply(subtract(12, 7), divide(2400, subtract(7, 3))) | subtract(n1,n0)|subtract(n2,n1)|divide(n3,#0)|multiply(#2,#1)| | general | B |
a fruit seller had some oranges . he sells 20 % oranges and still has 380 oranges . how many oranges he had originally ? | explanation : he sells 20 % of oranges and still there are 380 oranges remaining = > 80 % of oranges = 380 ⇒ ( 80 × total oranges ) / 100 = 380 ⇒ total oranges / 100 = 4.75 ⇒ total oranges = 4.75 × 100 = 475 answer : option a | a ) 475 , b ) 700 , c ) 220 , d ) 400 , e ) none of these | a | add(380, multiply(380, divide(20, const_100))) | divide(n0,const_100)|multiply(n1,#0)|add(n1,#1) | gain | A |
when positive integer n is divided by 3 , the remainder is 1 . when n is divided by 5 , the remainder is 3 . what is the smallest positive integer k such that k + n is a multiple of 15 ? | "n = 3 p + 1 = 5 q + 3 n + 2 = 3 p + 3 = 5 q + 5 n + 2 is a multiple of 3 and 5 , so it is a multiple of 15 . the answer is a ." | a ) 2 , b ) 4 , c ) 6 , d ) 8 , e ) 10 | a | subtract(15, reminder(3, 5)) | reminder(n3,n2)|subtract(n4,#0)| | general | A |
a square mirror has exactly half the area of the rectangular wall on which it is hung . if each side of the mirror is 34 inches and the width of the wall is 54 inches , what is the length of the wall , in inches ? | "since the mirror is 42 inches in all sides , it must be a square . area of a square is a = a ^ 2 ; 34 ^ 2 = 1156 . area of rectangle is double of that 2 * 1156 = 2312 . now a = lw and we need find w so a / l = w ; 2312 / 54 = 42.8 answer ! answer is c" | a ) 22.8 , b ) 33.8 , c ) 42.8 , d ) 63.8 , e ) 56.8 | c | divide(multiply(const_2, square_area(34)), 54) | square_area(n0)|multiply(#0,const_2)|divide(#1,n1)| | geometry | C |
in a certain parking lot , 10 % of the cars are towed for parking illegally . however 80 % of the cars which are parked illegally are not towed . what percentage of cars in the parking lot are parked illegally ? | "let x be the number of cars and let y be the number of cars parked illegally . 10 % * x = 20 % * y y / x = 1 / 2 = 50 % the answer is c ." | a ) 30 % , b ) 40 % , c ) 50 % , d ) 60 % , e ) 70 % | c | divide(subtract(const_1, divide(80, const_100)), divide(10, const_100)) | divide(n1,const_100)|divide(n0,const_100)|subtract(const_1,#0)|divide(#2,#1)| | gain | C |
a mixture of 125 gallons of wine and water contains 20 % of water . how much water must be added to the mixture in order to increase the percentage of water to 25 % of the new mixture ? | in 125 gallons of the solution there are 0.2 ∗ 125 = 25 gallons of water . we want to add w gallons of water to 125 gallons of solution so that 25 + w gallons of water to be 25 % of new solution : 25 + w = 0.25 ( 125 + w ) - - > w = 253 ≈ 8.33 . answer : d | a ) 10 gallons , b ) 8.5 gallons , c ) 8 gallons , d ) 8.33 gallons , e ) 9.33 gallons | d | divide(subtract(multiply(divide(multiply(divide(20, const_100), 125), const_100), 125), 25), subtract(const_1, divide(multiply(divide(20, const_100), 125), const_100))) | divide(n1,const_100)|multiply(n0,#0)|divide(#1,const_100)|multiply(n0,#2)|subtract(const_1,#2)|subtract(#3,n2)|divide(#5,#4) | general | D |
the volume of a rectangular swimming pool is 840 cubic meters and water is flowing into the swimming pool . if the surface level of the water is rising at the rate of 0.5 meters per minute , what is the rate s , in cubic meters per minutes , at which the water is flowing into the swimming pool ? | "the correct answer is e . there are not enough info to answer the question . a 840 cubic meters rectangle is built from : height * length * width . from the question we know the volume of the pool and the filling rate . a pool can have a height of 10 * width 8.4 * length 10 and have a volume of 840 cubic meters , and it can have a height of 1 meter , width of 100 meters and length of 8.4 . in both cases the pool will fill up in a different rate = e" | a ) 0.125 , b ) 0.25 , c ) 0.5 , d ) 0.75 , e ) not enough information to determine the rate | e | divide(840, 0.5) | divide(n0,n1)| | geometry | E |
gold is 19 times as heavy as water and copper is 9 times heavy . in what ratio must these metals be mixed so that the mixture may be 15 times as heavy as water ? | "? required ratio = 6 ? 4 = 3 : 2 answer b" | a ) 2 : 3 , b ) 3 : 2 , c ) 1 : 3 , d ) 2 : 1 , e ) none of these | b | divide(subtract(15, 9), subtract(19, 15)) | subtract(n2,n1)|subtract(n0,n2)|divide(#0,#1)| | general | B |
a vessel of capacity 2 litre has 25 % of alcohol and another vessel of capacity 6 litre had 50 % alcohol . the total liquid of 8 litre was poured out in a vessel of capacity 10 litre and thus the rest part of the vessel was filled with the water . what is the new concentration of mixture ? | "25 % of 2 litres = 0.5 litres 50 % of 6 litres = 3.0 litres therefore , total quantity of alcohol is 3.5 litres . this mixture is in a 10 litre vessel . hence , the concentration of alcohol in this 10 litre vessel is 35 % answer : e" | a ) 31 % . , b ) 71 % . , c ) 49 % . , d ) 29 % . , e ) 35 % . | e | multiply(divide(add(multiply(divide(25, const_100), 2), multiply(divide(50, const_100), 6)), 10), const_100) | divide(n1,const_100)|divide(n3,const_100)|multiply(n0,#0)|multiply(n2,#1)|add(#2,#3)|divide(#4,n5)|multiply(#5,const_100)| | general | E |
n is a whole number which when divided by 4 gives 3 as remainder . what will be the remainder when 2 * n is divided by 4 ? | explanation : let n = 4 * q + 3 . then , 2 * n = 8 * q + 6 = 4 ( 2 * q + 1 ) + 2 . thus when 2 * n is divided by 4 , the reminder is 2 . answer : b ) 2 | a ) 8 , b ) 2 , c ) 6 , d ) 7 , e ) 1 | b | reminder(multiply(3, 2), 4) | multiply(n1,n2)|reminder(#0,n0) | general | B |
how many pieces of 85 cm length can be cut from a rod of 25.5 meters long ? | "number of pieces = 2550 / 85 = 30 the answer is c ." | a ) 50 , b ) 40 , c ) 30 , d ) 20 , e ) 10 | c | divide(multiply(25.5, const_100), 85) | multiply(n1,const_100)|divide(#0,n0)| | physics | C |
when the number 9 y 30012 is exactly divisible by 11 , what is the smallest whole number that can replace y ? | "the given number = 9 y 30012 sum of the odd places = 2 + 0 + 3 + 9 = 14 sum of the even places = 1 + 0 + y ( sum of the odd places ) - ( sum of even places ) = number ( exactly divisible by 11 ) 14 - ( 1 + y ) = divisible by 11 13 � y = divisible by 11 . y must be 2 , to make given number divisible by 11 . b" | a ) 1 , b ) 2 , c ) 5 , d ) 6 , e ) 8 | b | subtract(9, 9) | subtract(n0,n0)| | general | B |
the dimensions of a rectangular solid are 6 inches , 5 inches , and 8 inches . if a cube , a side of which is equal to one of the dimensions of the rectangular solid , is placed entirely within thespherejust large enough to hold the cube , what the ratio of the volume of the cube to the volume within thespherethat is not occupied by the cube ? | answer : d . | ['a ) 10 : 17', 'b ) 2 : 5', 'c ) 5 : 16', 'd ) 10 : 19', 'e ) 32 : 25'] | d | divide(const_10, add(add(multiply(8, const_2), const_2), const_1)) | multiply(n2,const_2)|add(#0,const_2)|add(#1,const_1)|divide(const_10,#2) | geometry | D |
what is the unit digit in 4 ^ 100 ? | unit digit in 4 ^ 100 = unit digit in [ ( 4 ^ 4 ) ^ 25 ] but unit digit in 4 ^ 4 = 6 unit digit 4 ^ 100 = 6 answer is c | a ) 0 , b ) 1 , c ) 6 , d ) 3 , e ) 5 | c | reminder(power(4, const_2), const_10) | power(n0,const_2)|reminder(#0,const_10) | general | C |
a man walking at a rate of 15 km / hr crosses a bridge in 15 minutes . the length of the bridge is ? | "speed = 15 * 5 / 18 = 15 / 18 m / sec distance covered in 15 minutes = 15 / 18 * 15 * 60 = 750 m answer is d" | a ) 1250 m , b ) 1110 m , c ) 950 m , d ) 750 m , e ) 1300 m | d | multiply(divide(multiply(15, const_1000), const_60), 15) | multiply(n0,const_1000)|divide(#0,const_60)|multiply(n1,#1)| | gain | D |
( 26.3 * 12 * 20 ) / 3 + 125 = ? | b 2229 ? = [ ( 26.3 * 12 * 20 ) / 3 ] + 125 ? = 2104 + 125 = 2229 | a ) 2339 , b ) 2229 , c ) 4429 , d ) 5529 , e ) 2669 | b | add(divide(multiply(multiply(26.3, 12), 20), const_3), 125) | multiply(n0,n1)|multiply(n2,#0)|divide(#1,const_3)|add(n4,#2) | general | B |
among all sales staff at listco corporation , college graduates and those without college degrees are equally represented . each sales staff member is either a level - 1 or level - 2 employee . level - 1 college graduates account for 15 % of listco ' s sales staff . listco employs 60 level - 1 employees , 30 of whom are college graduates . how many sales staff members without college degrees are level - 2 employees ? | i ' m going in on this one . so let ' s say that we have the following so we know that l 1 = 60 and that c and l 1 = 0.15 x , we should set up a double set matrix btw but anyways , i ' m just explaining the point with this problem . now we are told that 0.15 x = 30 , therefore the grand total is 200 . now we know that l 2 is 200 - 60 = 140 . we also learn that c and no c are equally represented thus 100 each . therefore no c and no l 2 will be 100 - 30 = 70 . thus d is the correct answer choice | a ) 46 , b ) 42 , c ) 56 , d ) 70 , e ) 58 | d | divide(subtract(divide(30, divide(15, const_100)), 60), 2) | divide(n3,const_100)|divide(n6,#0)|subtract(#1,n4)|divide(#2,n1) | general | D |
one hour before john started walking from p to q , a distance of 23 miles , ann had started walking along the same road from q to p . ann walked at a constant speed of 3 miles per hour and john at 2 miles per hour . how many miles had ann walked when they met ? | ann walks from q to p at a speed of 3 miles / hr for one hour . she covers 3 miles in 1 hour and now distance between john and ann is 23 - 3 = 20 miles . ann walks at 3 mph and john at 2 mph so their relative speed is 3 + 2 = 5 mph . they have to cover 20 miles so it will take them 20 / 5 = 4 hours to meet . in 4 hrs , ann would have covered 4 hrs * 3 miles per hour = 12 miles . adding this to the 3 miles she covered before john , ann covered a total of 3 + 12 = 15 miles . answer ( e ) | a ) 6 miles , b ) 8,4 miles , c ) 9 miles , d ) 9,6 miles , e ) 15 miles | e | multiply(divide(23, add(3, 2)), 2) | add(n1,n2)|divide(n0,#0)|multiply(n2,#1) | physics | E |
a train 300 meters long completely crosses a 300 meters long bridge in 45 seconds . what is the speed of the train is ? | "s = ( 300 + 300 ) / 45 = 600 / 45 * 18 / 5 = 48 answer : b" | a ) 32 , b ) 48 , c ) 29 , d ) 27 , e ) 21 | b | divide(divide(add(300, 300), const_1000), divide(45, const_3600)) | add(n0,n1)|divide(n2,const_3600)|divide(#0,const_1000)|divide(#2,#1)| | physics | B |
a tank can be filled by two pipes a and b in 60 minutes and 40 minutes respectively . how many minutes will it take to fill the tank from empty state if b is used for the first half time and then a and b fill it together for the other half | "explanation : let the total time be x mins . part filled in first half means in x / 2 = 1 / 40 part filled in second half means in x / 2 = 1 / 60 + 1 / 40 = 1 / 24 total = x / 2 ∗ 1 / 40 + x / 2 ∗ 1 / 24 = 1 = > x / 2 ( 1 / 40 + 1 / 24 ) = 1 = > x / 2 ∗ 1 / 15 = 1 = > x = 30 mins option d" | a ) 15 mins , b ) 20 mins , c ) 25 mins , d ) 30 mins , e ) none of these | d | divide(multiply(multiply(40, const_3), const_2), add(const_4, const_4)) | add(const_4,const_4)|multiply(n1,const_3)|multiply(#1,const_2)|divide(#2,#0)| | physics | D |
a doctor prescribed 18 cubic centimeters of a certain drug to a patient whose body weight was 135 pounds . if the typical dosage is 2 cubic centimeters per 15 pounds of the body weight , by what percent was the prescribed dosage greater than the typical dosage ? | "typical dosage is dose : weight : : 2 : 15 . now if weight is 135 ( multiplying factor is 9 : ( 135 / 15 ) ) then typical dosage would be 2 * 9 = 18 cc . dosage = 18 cc . dosage is greater by 2 cc . % dosage is greater : ( 2 / 18 ) * 100 = 11.11 % c is the answer ." | a ) 8 % , b ) 9 % , c ) 11 % , d ) 12.5 % , e ) 14.8 % | c | multiply(divide(subtract(multiply(divide(2, 15), 135), 18), multiply(divide(2, 15), 135)), const_100) | divide(n2,n3)|multiply(n1,#0)|subtract(#1,n0)|divide(#2,#1)|multiply(#3,const_100)| | gain | C |
if the annual increase in the population of a town is 4 % and the present number of people is 15625 , what will the population be in 3 years ? | "the required population is = 15625 ( 1 + 4 / 100 ) ^ 3 = 15625 * 26 / 25 * 26 / 25 * 26 / 25 = 17576 answer is d" | a ) 15265 , b ) 16458 , c ) 25600 , d ) 17576 , e ) 14785 | d | multiply(multiply(divide(add(4, const_100), const_100), 15625), divide(add(4, const_100), const_100)) | add(n0,const_100)|divide(#0,const_100)|multiply(n1,#1)|multiply(#1,#2)| | gain | D |
john was thrice as old as tom 6 years ago . john will be 9 / 7 times as old as tom in 6 years . how old is tom today ? | "j - 6 = 3 ( t - 6 ) , so j = 3 t - 12 j + 6 = 9 / 7 * ( t + 6 ) 7 j + 42 = 9 t + 54 7 ( 3 t - 12 ) + 42 = 9 t + 54 12 t = 96 t = 8 the answer is b ." | a ) 6 , b ) 8 , c ) 10 , d ) 12 , e ) 14 | b | multiply(6, divide(9, 7)) | divide(n1,n2)|multiply(n0,#0)| | general | B |
a 300 m long train crosses a platform in 39 sec while it crosses a signal pole in 18 sec . what is the length of the platform ? | "speed = 300 / 18 = 50 / 3 m / sec . let the length of the platform be x meters . then , ( x + 300 ) / 39 = 50 / 3 3 x + 900 = 1950 = > x = 350 m . answer : b" | a ) 288 , b ) 350 , c ) 889 , d ) 276 , e ) 234 | b | subtract(multiply(speed(300, 18), 39), 300) | speed(n0,n2)|multiply(n1,#0)|subtract(#1,n0)| | physics | B |
a circular garden is surrounded by a fence of negligible width along the boundary . if the length of the fence is 1 / 6 of th area of the garden . what is the radius of the circular garden ? | "as per the question - - width is negligible now , let l be the length of the fence = 2 pir l = 1 / 6 ( pir ^ 2 ) pir ^ 2 = 12 pir r = 12 answer : e" | a ) 1 , b ) 2 , c ) 4 , d ) 8 , e ) 12 | e | multiply(const_2, sqrt(power(6, const_2))) | power(n1,const_2)|sqrt(#0)|multiply(#1,const_2)| | geometry | E |
mike drives his new corvette from san francisco to las vegas , a journey of 640 miles . he drives the first half of the trip at an average rate of 80 miles per hour , but has to slow down for the second half of his journey . if the second half of the trip takes him 200 percent longer than the first half , what is his average rate z in miles per hour for the entire trip ? | "veritas prepofficial solution correct answer : c using the formula : time = distance / rate , we find that mike takes 4 hours to cover the first 320 miles of his trip . since the 2 nd 320 miles take 200 % longer than the first , it takes mike 8 hours longer , or 12 hours . ( note : 200 % longer than the first half is not 200 % of the first half . ) the overall time is 4 hours + 12 hours or 16 hours . since the definition of average rate = total distance traveled / total time of travel , mike ' s average rate = 640 / 16 or 40 miles per hour . answer choice c is correct ." | a ) 26.7 , b ) z = 30.0 , c ) z = 40.0 , d ) z = 53.3 , e ) 60.0 | c | divide(640, add(add(multiply(divide(divide(640, const_2), 80), const_2), divide(divide(640, const_2), 80)), divide(divide(640, const_2), 80))) | divide(n0,const_2)|divide(#0,n1)|multiply(#1,const_2)|add(#1,#2)|add(#3,#1)|divide(n0,#4)| | physics | C |
a company organzied a help desk . there have been done 24 hrs work on software , 17 hrs on help user nd 40 % on other services . find the total no . of hrs have been worked on that day ? | let total no of hrs = x they have done 24 + 17 = 41 40 % of x is 40 / 100 = 2 / 5 x = 41 + 2 x / 5 x = 68.3 answer : a | a ) 68 , b ) 69 , c ) 70 , d ) 67 , e ) 76 | a | divide(multiply(add(const_1, const_4), add(24, 17)), const_3) | add(const_1,const_4)|add(n0,n1)|multiply(#0,#1)|divide(#2,const_3) | physics | A |
in an electric circuit , two resistors with resistances m and n are connected in parallel . in this case , if p is the combined resistance of these two resistors , then the reciprocal of p is equal to the sum of the reciprocals of m and n . what is p in terms of m and n ? | the wording is a bit confusing , though basically we are told that 1 / p = 1 / m + 1 / n , from which it follows that p = mn / ( m + n ) . answer : b | ['a ) ( n - m )', 'b ) mn / ( m + n )', 'c ) ( nm )', 'd ) ( n - m ) / ( m + n )', 'e ) none of these'] | b | divide(multiply(const_1, const_3), add(const_1, const_3)) | add(const_1,const_3)|multiply(const_1,const_3)|divide(#1,#0) | geometry | B |
the sum of money will be double itself in 10 years and simple interest find rate of interest ? | t = 10 years p = principle amount = x a = total amount = 2 x si = simple interest = a - p = 2 x - x = x r = 100 si / pt = 100 x / 10 x = 10 % answer is a | a ) 10 % , b ) 15 % , c ) 20 % , d ) 25 % , e ) 30 % | a | multiply(divide(10, const_100), const_100) | divide(n0,const_100)|multiply(#0,const_100) | gain | A |
the front wheels of a wagon are 2 π feet in circumference and the rear wheels are 3 π feet in circumference . when the front wheels have made 10 more revolutions than the rear wheels , how many feet has the wagon travelled ? | solution let the rear wheel make x revolutions . then , the front wheel makes ( x + 10 ) revolutions . ( x + 10 ) x 3 π = x × 2 π ‹ = › 3 x + 30 = 2 x ‹ = › x = 30 . distance travelled by the wagon = ( 2 π x 30 ) ft ‹ = › ( 60 π ) ft . answer c | a ) 30 π , b ) 45 π , c ) 60 π , d ) 90 π , e ) none | c | divide(multiply(multiply(multiply(2, 10), 3), multiply(subtract(const_12, const_1), 2)), add(const_3, const_4)) | add(const_3,const_4)|multiply(n0,n2)|subtract(const_12,const_1)|multiply(n1,#1)|multiply(n0,#2)|multiply(#3,#4)|divide(#5,#0) | geometry | C |
the speed of a subway train is represented by the equation z = s ^ 2 + 2 s for all situations where 0 ≤ s ≤ 7 , where z is the rate of speed in kilometers per hour and s is the time in seconds from the moment the train starts moving . in kilometers per hour , how much faster is the subway train moving after 7 seconds than it was moving after 2 seconds ? | "given : z = s ^ 2 + 2 s for 0 ≤ s ≤ 7 z ( 2 ) = 2 ^ 2 + 2 * 2 = 8 z ( 7 ) = 7 ^ 2 + 2 * 7 = 63 therefore z ( 7 ) - z ( 3 ) = 63 - 8 = 55 km / hr option b" | a ) 4 , b ) 55 , c ) 15 , d ) 48 , e ) 63 | b | subtract(add(power(7, 2), multiply(7, 2)), add(power(2, 2), multiply(2, 2))) | multiply(n0,n4)|multiply(n0,n5)|power(n4,n0)|power(n5,n0)|add(#0,#2)|add(#1,#3)|subtract(#4,#5)| | physics | B |
a fair price shopkeeper takes 10 % profit on his goods . he lost 50 % goods during theft . his loss percent is : | "explanation : suppose he has 100 items . let c . p . of each item be re . 1 . total cost = rs . 100 . number of items left after theft = 50 . s . p . of each item = rs . 1.10 total sale = 1.10 * 50 = rs . 55 hence , loss % = 45 / 100 * 100 = 45 % answer : b" | a ) 72 % , b ) 45 % , c ) 32 % , d ) 12 % , e ) 22 % | b | subtract(const_100, subtract(add(const_100, 10), divide(multiply(add(const_100, 10), 50), const_100))) | add(n0,const_100)|multiply(n1,#0)|divide(#1,const_100)|subtract(#0,#2)|subtract(const_100,#3)| | gain | B |
how many liters of a 40 % iodine solution need to be mixed with 35 liters of a 20 % iodine solution to create a 30 % iodine solution ? | "solution 1 : assume the iodine solution to be mixed = x lts . iodine = 0.4 x lts , water = 0.6 x lts . solution 2 : 35 liters of a 20 % iodine solution iodine = 7 lts , water = 28 lts . total iodine = 0.4 x + 7 total water = 0.6 x + 28 the resultant is a 35 % idoine solution . hence ( 0.4 x + 7 ) / ( x + 35 ) = 30 / 100 40 x + 700 = 30 x + 1050 10 x = 305 x = 30.5 lts correct option : a" | a ) 30.5 , b ) 49 , c ) 100 , d ) 105 , e ) 140 | a | add(divide(subtract(multiply(divide(multiply(35, 20), const_100), 30), multiply(divide(multiply(35, 20), const_100), 20)), subtract(multiply(20, divide(40, const_100)), divide(multiply(35, 20), const_100))), 35) | divide(n0,const_100)|multiply(n1,n2)|divide(#1,const_100)|multiply(n2,#0)|multiply(n3,#2)|multiply(n2,#2)|subtract(#3,#2)|subtract(#4,#5)|divide(#7,#6)|add(n1,#8)| | gain | A |
cole drove from home to work at an average speed of 75 kmh . he then returned home at an average speed of 105 kmh . if the round trip took a total of 6 hours , how many minutes did it take cole to drive to work ? | "first round distance travelled ( say ) = d speed = 75 k / h time taken , t 2 = d / 75 hr second round distance traveled = d ( same distance ) speed = 105 k / h time taken , t 2 = d / 105 hr total time taken = 6 hrs therefore , 6 = d / 75 + d / 105 lcm of 75 and 105 = 525 6 = d / 75 + d / 105 = > 6 = 7 d / 525 + 5 d / 525 = > d = 525 / 2 km therefore , t 1 = d / 75 = > t 1 = 525 / ( 2 x 75 ) = > t 1 = ( 7 x 60 ) / 2 - - in minutes = > t 1 = 210 minutes . d" | a ) 84 , b ) 136 , c ) 172 , d ) 210 , e ) 478 | d | multiply(divide(multiply(105, 6), add(75, 105)), const_60) | add(n0,n1)|multiply(n1,n2)|divide(#1,#0)|multiply(#2,const_60)| | physics | D |
a salesman ’ s terms were changed from a flat commission of 5 % on all his sales to a fixed salary of rs . 1400 plus 2.5 % commission on all sales exceeding rs . 4,000 . if his remuneration as per new scheme was rs . 600 more than that by the previous schema , his sales were worth ? | "[ 1400 + ( x - 4000 ) * ( 2.5 / 100 ) ] - x * ( 5 / 100 ) = 600 x = 20000 answer : c" | a ) 12028 , b ) 12000 , c ) 20000 , d ) 12197 , e ) 12012 | c | divide(600, divide(5, const_100)) | divide(n0,const_100)|divide(n4,#0)| | general | C |
the sum of four consecutive even numbers is 52 , find the third number ? | "let the first number be n let the second number be n + 2 let the third number be n + 4 let the third number be n + 6 n + n + 2 + n + 4 + n + 6 = 52 combine like terms : 4 n + 12 = 52 4 n + 12 = 52 4 n = 52 - 12 4 n = 40 n = 10 n + 2 = 12 n + 4 = 14 n + 6 = 16 the third number is 14 answer : a" | a ) 14 , b ) 10 , c ) 12 , d ) 16 , e ) 20 | a | add(add(power(add(add(divide(subtract(subtract(52, const_10), const_2), const_4), const_2), const_2), const_2), power(add(add(add(divide(subtract(subtract(52, const_10), const_2), const_4), const_2), const_2), const_2), const_2)), add(power(divide(subtract(subtract(52, const_10), const_2), const_4), const_2), power(add(divide(subtract(subtract(52, const_10), const_2), const_4), const_2), const_2))) | subtract(n0,const_10)|subtract(#0,const_2)|divide(#1,const_4)|add(#2,const_2)|power(#2,const_2)|add(#3,const_2)|power(#3,const_2)|add(#5,const_2)|add(#4,#6)|power(#5,const_2)|power(#7,const_2)|add(#9,#10)|add(#11,#8)| | physics | A |
a mixture contains alcohol and water in the ratio 4 : 3 . if 5 liters of water is added to the mixture , the ratio becomes 4 : 5 . find the quality of alcohol in the given mixture . | "let the quantity of alcohol and water be 4 x and 3 x 4 x / ( 3 x + 5 ) = 4 / 5 20 x = 4 ( 3 x + 5 ) x = 2.5 quantity of alcohol = 4 * 2.5 = 10 liters . answer is a" | a ) 10 , b ) 8 , c ) 7.5 , d ) 6 , e ) 3.5 | a | multiply(5, const_1) | multiply(n2,const_1)| | general | A |
two numbers are respectively 71 % and 80 % more than a third number . the percentage that is first of the second is ? | "i ii iii 171 180 100 180 - - - - - - - - - - 171 100 - - - - - - - - - - - ? = > 5 % answer : b" | a ) 4 % , b ) 5 % , c ) 6 % , d ) 7 % , e ) 8 % | b | subtract(const_100, multiply(divide(add(71, const_100), add(80, const_100)), const_100)) | add(n0,const_100)|add(n1,const_100)|divide(#0,#1)|multiply(#2,const_100)|subtract(const_100,#3)| | gain | B |
the sum of the first 50 positive even integers is 2550 . what is the sum w of even integers from 102 to 200 inclusive ? | "my solution is : first 50 even integers : 2 4 6 8 < . . . > integers from 102 to 200 102 104 106 108 < . . . > we notice that each integer from the second set is 100 more than the respective integer in the first set . since we have 50 even integers from 102 to 200 , then : w = 2550 + ( 100 * 50 ) = 7550 . b" | a ) 5100 , b ) 7550 , c ) 10100 , d ) 15500 , e ) 20100 | b | multiply(divide(add(200, 102), const_2), add(divide(subtract(200, 102), const_2), const_1)) | add(n2,n3)|subtract(n3,n2)|divide(#1,const_2)|divide(#0,const_2)|add(#2,const_1)|multiply(#4,#3)| | general | B |
| 16 - 5 | - | 5 - 12 | = ? | "| 16 - 5 | - | 5 - 12 | = | 11 | - | - 7 | = 11 - 7 = 4 correct answer e" | a ) 3 , b ) 2 , c ) 1 , d ) 0 , e ) 4 | e | subtract(subtract(16, 5), subtract(12, 5)) | subtract(n0,n1)|subtract(n3,n2)|subtract(#0,#1)| | general | E |
in a certain brick wall , each row of bricks above the bottom row contains one less brick than the row just below it . if there are 5 rows in all and a total of 200 bricks in the wall , how many bricks does the bottom row contain ? | "the bottom row has x bricks x + x - 1 + x - 2 + x - 3 + x - 4 = 200 5 x - 10 = 200 5 x = 190 x = 38 answer : e" | a ) 42 , b ) 35 , c ) 40 , d ) 33 , e ) 38 | e | divide(subtract(subtract(subtract(subtract(200, const_1), const_2), const_3), const_4), 5) | subtract(n1,const_1)|subtract(#0,const_2)|subtract(#1,const_3)|subtract(#2,const_4)|divide(#3,n0)| | general | E |
if m and n are positive integer , and 1800 m = n ^ 3 , what is the least possible value of m ? | 1800 * m = n ^ 3 m = n ^ 3 / 1800 = n ^ 3 / ( 3 ^ 2 * 5 ^ 2 * 2 ^ 3 ) so n has to be 3 * 5 * 2 so that n ^ 3 is divisible by 1800 . so m = ( 2 * 3 * 5 ) ^ 3 / ( 3 ^ 2 * 5 ^ 2 * 2 ^ 3 ) m = 15 answer : c | a ) 2 , b ) 3 , c ) 15 , d ) 30 , e ) 45 | c | floor(power(multiply(1800, const_2), const_0_33)) | multiply(n0,const_2)|power(#0,const_0_33)|floor(#1) | general | C |
find the value of a / b + b / a , if a and b are the roots of the quadratic equation x 2 + 4 x + 2 = 0 ? | "a / b + b / a = ( a 2 + b 2 ) / ab = ( a 2 + b 2 + a + b ) / ab = [ ( a + b ) 2 - 2 ab ] / ab a + b = - 4 / 1 = - 4 ab = 2 / 1 = 2 hence a / b + b / a = [ ( - 4 ) 2 - 2 ( 2 ) ] / 2 = 4 / 2 = 2 . c )" | a ) 8 , b ) 10 , c ) 2 , d ) 16 , e ) 24 | c | subtract(divide(power(negate(4), 2), 2), 2) | negate(n1)|power(#0,n0)|divide(#1,n2)|subtract(#2,n0)| | general | C |
a man sitting in a train which is traveling at 80 kmph observes that a goods train , traveling in opposite direction , takes 9 seconds to pass him . if the goods train is 280 m long , find its speed . ? | "relative speed = 280 / 9 m / sec = ( ( 280 / 9 ) * ( 18 / 5 ) ) kmph = 112 kmph . speed of goods train = ( 112 - 80 ) kmph = 32 kmph . answer : a" | a ) 32 kmph , b ) 58 kmph , c ) 62 kmph , d ) 65 kmph , e ) 75 kmph | a | subtract(multiply(divide(280, 9), const_3_6), 80) | divide(n2,n1)|multiply(#0,const_3_6)|subtract(#1,n0)| | physics | A |
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